MTH5104: Convergence and Continuity 2016/2017
Coursework 3 – SOLUTIONS
Warm-Up Questions.
1. (a) The sequence (xn )∞
n=1 given by xn =
1
n+10 cos(πn)
does converge to zero.
Proof. Given ε > 0, let N = d1/εe + 10 (so 1/(N − 10) ≤ ε). Then for
all n > N we have
|xn | =
1
n+10 cos(πn)
≤
1
n−10
<
1
N −10
≤ε
(since −1 ≤ cos(πn) ≤ 1).
(b) The sequence (xn )∞
n=1 given by xn = 1 if n is a perfect square and xn = 0
otherwise does not converge to zero.
Proof. Let ε = 0.5. Then given any natural number N , let n = (N + 1)2 .
Now |xn | = 1 ≥ ε since n is a perfect square.
√
n−
√
n − 1. In particular, all xn are positive.
√
√
(a) Using (a−b)(a+b)√= a2 −b2 , we have xn ·( n+ n − 1) = n−(n−1) = 1
√
√
so xn = 1/( n + n − 1) ≤ 1/ n.
√
(b) Let us first prove that the sequence (yn ) given by yn = 1/ n converges
to zero.
√
Proof. Given ε > 0, let N = d1/ε2 e (so 1/N ≤ ε2 or equivalently 1/ N ≤
ε). Then for all n > N we have
2. We have xn =
|yn | =
√1
n
<
√1
N
≤ ε.
By the above |xn | = xn ≤ yn = |yn |, so by Lemma 2.3 the sequence (xn )
also converges to zero.
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Convergence and Continuity 2016/2017
Coursework 3 Solutions
Feedback Question.
3. (a) The sequence (xn )∞
n=1 given by xn = 5/n converges to zero.
Proof. Given ε > 0, let N = d5/εe (so N ≥ 5/ε, so 5/N ≤ ε). Now
∀n > N we have |xn | = 5/n < 5/N ≤ ε.
This is equivalent to the following winning strategy for the demon game:
Suppose the demon picks ε > 0.
We pick N = d5/εe (so N ≥ 5/ε, so 5/N ≤ ε).
Then suppose the demon picks n > N .
Then |xn | = 5/n < 5/N ≤ ε so we win.
(b) The sequence (xn )∞
n=1 given by xn =
(−1)n
200
does not converge to zero.
Proof. Let ε = 1/300. Then given any natural number N let n = N + 1.
So |xn | = 1/200 ≥ ε.
This proof is equivalent to the following winning strategy for the negated
demon game:
We pick ε = 1/300.
Then suppose the demon picks N .
We pick n = N + 1.
Then |xn | = 1/200 ≥ ε so we win (the negated game).
(c) The sequence (xn )∞
n=1 given by xn =
1
n!
converges to zero.
Proof. Given ε > 0, let N = d1/εe (so N ≥ 1/ε, so 1/N ≤ ε). Now
∀n > N we have |xn | = 1/n! ≤ 1/n < 1/N ≤ ε.
Equivalent winning strategy for demon game:
Suppose the demon picks ε > 0.
We pick N = d1/εe (so N ≥ 1/ε, so 1/N ≤ ε).
Then suppose the demon picks n.
Now |xn | = 1/n! ≤ 1/n < 1/N ≤ ε, so we win.
(d) The sequence (xn )∞
n=1 given by xn =
√1
n+1
converges to zero.
√
2
2 , so 1/ N ≤ ε). Now
Proof. Given ε > 0, let N =
≥ 1/ε√
√ d1/ε e (so N
√
∀n > N we have |xn | = 1/( n + 1) < 1/ n < 1/ N ≤ ε.
Equivalent winning strategy for demon game:
Suppose the demon picks ε > 0.
√
We pick N = d1/ε2 e (so N ≥ 1/ε2 , so 1/ N ≤ ε).
Then suppose the
n.
√
√ demon picks
√
Now |xn | = 1/( n + 1) < 1/ n < 1/ N ≤ ε, so we win.
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Convergence and Continuity 2016/2017
Coursework 3 Solutions
1
−8
converges to zero.
(e) The sequence (xn )∞
n=1 given by xn = n
√
1
Proof. Given ε > 0, let N = d1/ε8 e (so N ≥ 1/ε8 , so N − 8 = 1/ 8 N ≤ ε).
1
1
Now ∀n > N we have |xn | = n− 8 < N − 8 ≤ ε.
Equivalent winning strategy for demon game:
Suppose the demon picks ε > 0.
√
1
We pick N = d1/ε8 e (so N ≥ 1/ε8 , so N − 8 = 1/ 8 N ≤ ε).
Then suppose the demon picks n.
1
1
Now |xn | = n− 8 < N − 8 ≤ ε, so we win.
Extra Questions.
4. If you could not solve this, study the solution to the first part and then try to
go back and solve the other parts in a similar fashion!
As (xn ) converges to zero, we know (by definition) that for every ε > 0 there
exists some Nx ∈ N such that ∀n > Nx : |xn | < ε. We now claim that (x̃n )
converges to zero.
Proof. Given ε > 0, let N = max{Nx , 10}. Then ∀n > N we have n > 10 so
x̃n = xn . Moreover, we have n > Nx , so |x̃n | = |xn | < ε.
As (yn ) converges to zero, we know (by definition) that for every ε > 0 there
exists some Ny ∈ N such that ∀n > Ny : |yn | < ε. We now claim that (ỹn )
converges to zero.
Proof. Given ε > 0, let N = max{Ny , 1000}. Then ∀n > N we have n > 1000
so ỹn = yn . Moreover, we have n > Ny , so |ỹn | = |yn | < ε.
Remark. Note that we don’t even see in the proof to which value we changed
the first 10 (or 1000) elements of the sequence.
As (zn ) does not converge to zero, we know that there exists ε > 0 such that
∀N ∈ N ∃n > N : |zn | ≥ ε. We now claim that (z̃n ) does not converge to
zero.
Proof. First, we pick the same ε > 0 as above. Then, given any N ∈ N (by
the Demon) we set Ñ = max{N, 1010 }. From the fact that (yn ) does not
converge to zero, if we would have been given this Ñ by the Demon, we could
have found some n0 > Ñ such that |yn0 | ≥ ε. We pick exactly this n0 . As
n0 > Ñ ≥ N , this is allowed. Moreover, as n0 > Ñ ≥ 1010 , we have ỹn0 = yn0 ,
so |ỹn0 | = |yn0 | ≥ ε.
Reto Buzano, 8 October 2016
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